Pre-Calc Lesson 5.1 Exponents -- Growth and Decay: Integral Exponents

Laws of Exponents:

1. bxby = bx+y 2. bx = bx-y 3. If b /= 0,1,or -1, by then bx = by iff x = y

1. (ab)x = axbx 5. (a)x = ax

(b) bx

1. If x ≠ 0, and a>0, and b>0, then ax = bx iff a = b

1. (bx)y = bxy 8. b0 = 1 9. b-x = 1 bx

Example 1: Simplify: (b2/a)-2 and (a3/b)-3

1st – the negative on the exponent indicates taking the reciprocal of the base! soo-- (a/b2)2 and (b/a3)3Now distribute the exponent on the outside of the parentheses a1x2 ; b1x3

b2x2 a3x3

voila -- a2 ; b3

b4 a9

Example 2: Simplify: (a2 + b2)-1 where a ≠ 0 and b ≠ 01st – make the inside of the parentheses look like one big fraction. – (a2 + b2)-1

1Now take the reciprocal of the whole thing . ( 1 )1

a2 + b2

Since the power is 1 this is not needed soo --- 1 a2 + b2

Example 3: Simplify: a) x5 + x-2 and b) (x5)(x-2) x-3 x-3

•x5 + (1/x2) (1/x3)Now multiply top and bottom by the LCD -- x3

x3(x5 + (1/x2) x3 (1/x3) x8 + x = x8 + x 1

•Much different problem as there is no addition or subtraction -- x5 + (-2) – (-3)

= x5 – 2 + 3

= x6

Suppose that the cost of a hamburger has been increasing at the rate of 9% per year. Then each year the cost is 1.09 times the cost the previous year. Suppose that the cost now is $4.

Some projected future costs would be:

Time (years from now) | 0 | 1 | 2 | 3 | ‘t’ |Cost (dollars) | 4 | 4(1.09)| ? | ? | ? | 4(1.09)2 4(1.09)3 4(1.09)t The above table suggest that the cost is a function of time ‘t’. Since the variable ‘t’ occurs as an exponent, the cost is said to be an exponential function of time ‘t’.

C(t) = ??????? 4(1.09)t

Example 4: Using the cost function described earlier, find the cost of a hamburger:

a) 5 years from now. 4(1.09)5 = $6.15

b) 5 years ago. 4(1.09)- 5 = $2.60

Look at the graphs of the two functions at the top of page170, can you tell which graph describes ‘exponential growth’ and which graph describes ‘exponential decay’ ?

Generally, all exponential growth or decay problems can be described by this model:

A(t) = A0(1 + r)t where A0 the initial amount (or when ‘t’ = 0) r the ‘growth’ rate (usually a percent) t the time (usually in terms of years)

Example 5: Suppose that a radioactive isotope decays so that the radioactivity present decreases 15% per day.

If 40 g are present now, find the amount present:

a) 6 days from now b) 6 days ago A(6) = A(-6) = = 40(1 + (-.15))6 = 40(1 + (-.15)) – 6

= 15.09 = 106.06Hw: pg 172 CE: #1-17 all; WE: #1-19 odd